The n-Category Café

Thanks — I’ll add that link to TWF.

David wrote:

Why is the Euler characteristic defined as an alternating sum?

There are many answers to this question, some going back to Euler. But here’s one: the Euler characteristic $\chi$, defined as an alternating sum of ranks of homology groups, has the marvelous property that

$$\chi(E) = \chi(F) \chi(B)$$

for any fibration

$$F \to E \to B$$

of connected spaces with finite Euler characteristic.

In other words, it’s multiplicative not just for ordinary products but also for ‘twisted’ products: the total space $E$ is like a ‘twisted product’ of the base space $B$ and the fiber $F$.

(If that doesn’t make sense, imagine a Möbius strip.)

Similarly, the homotopy cardinality $| \cdot |$, defined as the alternating product of cardinalities of homotopy groups, has

$$|E| = |F| |B|$$

for any fibration

$$F \to E \to B$$

of connected spaces with finite homotopy cardinality.

The fun part is to see why clever additive cancellations make this fact true for the Euler characteristic, while clever multiplicative cancellations make this fact true for the homotopy cardinality!

Is it because it’s dealing with the exponents of the nontorsion part of homology, i.e., the ranks?

The ranks of finitely generated abelian groups add when we multiply those groups:

$$rk(A \times B) = rk(A) + rk(B)$$

So, there’s some sort of logarithmic thing going on when we measure the size of an abelian group by taking its rank. But of course the multiplicative-additive distinction is a bit weird here, because $A \times B = A + B$ in the category of abelian groups!

So why does one generally exclude the torsion part?

The traditional reason is that lots of stuff works fine even if we ignore it.

But, that might not be the best approach. Dan Christensen and were thinking about this a bit. You could try working with the cohomology with coefficients in $\mathbb{Z}/p$ and get some sort of ‘mod $p$ Euler characteristic’ for each prime $p$, or maybe each prime power, and then maybe try to fit these together into some sort of ‘adelic’ Euler characteristic that keeps track of all primes at once.

I have a feeling that some people know a lot more about this than I do! But I don’t know who they are.

Could your general conjecture about Euler characteristic and homotopy cardinality suggest interesting facts?

I mean, imagine we hadn’t calculated the homotopy groups of the spheres beyond $\pi_n(S^n) = \mathbb{Z}$. Wouldn’t we know that because $\chi(S^{2m}) = 2$ that another $\mathbb{Z}$ would have to appear for at least one $(2k + 1)^{th}$ group, $\pi_{2k + 1}(S^{2m})$.

Or is it that the kind of erratic alternating product we’re dealing with can overcome a $\mathbb{Z}$ just using finite groups?

I seem to remember thinking this through before. Isn’t it just that the only difference between the alternating product for $S^3$ and for $S^2$ is the former’s 1 where the latter scores a $|\mathbb{Z}|$? So, the homotopy cardinality of $S^3$ would be $2 / \infty = 0$, matching the Euler characteristic.

David wrote:

I seem to remember thinking this through before. Isn’t it just that the only difference between the alternating product for $S^3$ and for $S^2$ is the former’s 1 where the latter scores a $|\mathbb{Z}|$? So, the homotopy cardinality of $S^3$ would be $2 / \infty = 0$, matching the Euler characteristic.

Right! So, my challenge was sort of a joke. I asked “if I can show that some divergent product equals 2, can you show that dividing it by $\infty$ gives 0?” And the joke is that I could have made this question much harder, just by turning it around: “if I can how that some divergent product equals 0, can you show that multiplying it by $\infty$ gives 2?”

When you thought this through before, had you known the homotopy groups of $S^3$ mostly matched those of $S^2$? For some reason I was shocked to discover this. It’s like discovering you’ve never noticed that two good friends of yours are twins!

If you think that’s cool, you should see it in pictures.

A general position immersed curve (which may have multiple components) represents a 1-morphism from the null object to the null object. Such morphisms are generated by |,X, cup, and cap. Generating 2-morphisms are given by a birth, death, saddle, [| =>zig-zag] and reverse, a psi as above, the projection of a type II move, and the projection of a type III move.

A general position immersed sphere (or any surface) in 3-space can be arranged so that a sequence of horizontal planes chosen to flank all critical (in a general sense) phenomena. The intersection of such a plane with the sphere is an immersed curve or a 1-morphism [ null->null]. The transition from one cross section to another is given by one of the above generating 2-morphisms (tensored and composed with identities on the unaffected regions). So each immersed sphere represents a composition of 2-morphisms

[Null => [null –0–> null] => … => [null –0–> null] => Null]

where Null is the empty 2-morphism from the empty object to itself and [null –0–> null] is the 1-morphism that corresponds to a simple convex closed curve [CIRCLE].

The changes between spheres in the process can also be arranged to involve generic critical points. And these local changes are generated by the movie-moves that don’t involve type I moves (aka branch points). Such a move between immersed spheres, then, can be thought of as a 3-morphism. Let +0 denote a counter clockwise oriented circle, and -0 denote a clockwise oriented circle.
A sphere eversion is the 3-morphism

[Null => +0 => Null] ===> [Null => -0 => Null].

You can cap off one end with the 3-morphisms that start from the NULL 3-morphism and go to an embedded 2-sphere, then cap off the opposite end with the opposite 3-morphism. The result is a generator of the 3rd stable stem.

I tried that link
but didn’t see any polyhedral model
??

Right — this is the Sloane’s entry that I mentioned in this Week’s Finds.

Why do you mention Batanin’s in particular?

As you know, a category enriched over some category $K$ is one where instead of hom-sets $hom(x,y)$, we have hom-objects $hom(x,y) \in K$. Tom defines a concept of cardinality enriched over $K$ when $K$ is a category whose objects are already equipped with a concept of cardinality. Since strict $(n+1)$-categories are categories enriched over $n Cat$, and everyone knows the cardinality of an $0$-category — since that’s just a set — by induction we get a concept of cardinality for strict $n$-categories!

Batanin’s weak $n$-categories are close to the usual strict ones in the following sense: for any pair of objects $x,y$ you get an $(n-1)$-category $hom(x,y)$. This is similar to the concept of enrichment — though it’s not so simple, since the usual category axioms hold only ‘weakly’.

So, I can vaguely imagine doing an inductive definition of cardinality for Batanin’s weak $n$-categories that mimics what Tom did.

However, it was just a vague thought. It could be wrong: it could be that Trimble’s definition makes things easier, since it’s more closely based on the concept of ‘iterated enrichment’.

As for uses, perhaps we could employ it profitably on those fundamental $n$-categories of stratified spaces we looked at once.